Question

need help for integration problem ƒdx/1+√x

Answer 1

Is that the
integral of the derivative of 1 + x^1/2 ? ?

Answer 2

Because the integral and derivative cancel each other out, leaving 1 + x^1/2
What are the limits of the integral?

Answer 3

1/1 + x^1/2

Answer 4

integral is undefineid

Answer 5

\[\int\limits_{?}^{?} 1 + x ^{1/2}\]

Is that it?

Answer 6

Then the answer is jut 1 + x^1/2

Answer 7

\[\int\limits_{?}^{?}1\div1+\]

Answer 8

I think he means$$\int\frac1{1+\sqrt x}dx$$

Answer 9

yes man

Answer 10

Lol wow thanks oldrin.

Answer 11

in which case you multiply top and bottom by \(1-\sqrt{x}\) to get:$$\int \frac{1-\sqrt{x}}{1-x}dx=\int\frac1{1-x}dx-\int\frac{\sqrt{x}}{1-x}dx$$

Answer 12

can ı do this for indefineid integral because

Answer 13

maybe x=0

Answer 14

the first integral is trivial:$$\int\frac1{1-x}dx=-\log(1-x)+C$$

Answer 15

the second integral is easy once we let \(u=\sqrt{x}\):$$2u\ du=dx$$so we get$$\int\frac{u}{1-u^2}\cdot2u\ du=2\int\frac{u^2}{1-u^2}du=2\int\left(\frac1{1-u^2}-1\right)du$$

Answer 16

by partial fractions we have that$$\frac1{1-u^2}=\frac{A}{1+u}+\frac{B}{1-u}$$where \(A(1-u)+B(1+u)=1\) so we see that \(A+B=1\) and \(-A+B=0\) so that \(A=B=1/2\) and we have:$$\begin{align*}\int\left(\frac1{1-u^2}-1\right)du&=\int\left(\frac12\cdot\frac1{1+u}+\frac12\cdot\frac1{1-u}-1\right)du\\&=\frac12\log(1+u)-\frac12\log(1-u)-u+C\\&=\frac12\log(1+\sqrt{x})-\frac12\log(1-\sqrt{x})-\sqrt{x}+C\end{align*}$$

Answer 17

so our final result is $$-\log(1-x)+\log(1+\sqrt{x})-\log(1-\sqrt{x})-2\sqrt{x}+C$$

Answer 18

hmm nah there msut be something wrong there

Answer 19

I confused a little :D

Answer 20

\[\int\limits_{}^{}\frac{1}{1+\sqrt{x}} dx\]
Or we could have just started with the sub \[u=\sqrt{x} => u^2=x => 2u du=dx\]
\[\int\limits_{}^{}\frac{2u du}{1+u} \]
The maybe do another sub like v=1+u so u=v-1 and dv=du
\[2 \int\limits_{}^{}\frac{v-1}{v} dv\]
You should be able to take it pretty easy for here

Answer 21

lool that works too

Answer 22

the error in my approach was a small sign screw up:$$-\log(1-x)-\log(1+\sqrt{x})+\log(1-\sqrt{x})+2\sqrt{x}+C$$

Answer 23

oh its very easy

Answer 24

thanks for everyone

Answer 25

can we use two sub in this question

Answer 26

I did.

Answer 27

See above.

Answer 28

yes ı saw thanks a lot

Answer 29

You can do multiple subs for any problem

Answer 30

yes can ı ask you a question

Answer 31

You could have wrote the two subs as one sub those

Answer 32

\[\int\limits_{}^{}\frac{1}{1+\sqrt{x}} dx\]
Let \[v=1+\sqrt{x} =>v-1=\sqrt{x} => (v-1)^2=x => 2(v-1) dv=dx\]

Answer 33

\[\int\limits_{}^{}\frac{2(v-1)dv}{v}\]

Answer 34

Go ahead, what is the question?

Answer 35

could we here with the multiplication of the expression x or conjugate for example \[\sqrt{x}-1 \]

Answer 36

You mean multiply the conjugate of sqrt(x)+1 on bottom and top?

I think that is what the other dude did as a first step.

Answer 37

Why do that when you just need a simple sub though

Answer 38

yes ı mean this is it possible

Answer 39

the answer is \[2\sqrt{x}+2-2\ln \left| \sqrt{x}+1 \right| + c\] isnt this